baye's rule 1. Suppose a certain drug test is 99% accurate, that is, the test will correctly identify a drug user as testing positive 99% of the time, and will correctly identify a non-user as testing negative 99% of the time.Let's assume a corporation decides to test its employees for opium use, and 0.5% of the employees use the drug. We want to know the probability that, given a positive drug test, an employee is actually a drug user. Let "D" be the event of being a drug user and "N" indicate being a non-user. Let "+" be the event of a positive drug test. We need to know the following:

Pr(D), or the probability that the employee is a drug user, regardless of any other information. This is 0.005, since 0.5% of the employees are drug users.

Pr(N), or the probability that the employee is not a drug user. This is 1-Pr(D), or 0.995.

Pr(+|D), or the probability that the test is positive, given that the employee is a drug user. This is 0.99, since the test is 99% accurate.

Pr(+|N), or the probability that the test is positive, given that the employee is not a drug user. This is 0.01, since the test will produce a false positive for 1% of non-users.

Pr(+), or the probability of a positive test event, regardless of other information. This is 0.015 or 1.5%, which found by adding the probability that the test will produce a true positive result in the event of drug use (= 99% x 0.5% = 0.495%) plus the probability that the test will produce a false positive in the event of non-drug use (= 1% x 99.5% = 0.995%).

sol. = 0.99x0.005 0.99x0.005+ 0.01x0.995 =0.3322

2. At a certain university, 4% of men are over 6 feet tall and 1% of women areover 6 feet tall. The total student population is divided in the ratio 3:2 infavour of women. If a student is selected at random from among all those oversix feet tall, what is the probability that the student is a woman?

Let M={Student is Male}, F={Student is Female}, (note that M and F partition the samplespace of students), T={Student is over 6 feet tall}. We know that P(M) = 2/5, P(F) = 3/5,P(T|M) = 4/100 and P(T|F) = 1/100. We require P(F|T). Using Bayes’ Theorem we have: P(F\T) = P(T\F) P(F) P(T\F)P(F)+P(T\M)P(M) = 1/100 x 3/5 1/100 x 3/5 x 4/ 100 x 2/5 = 3/11

3. A factory production line is manufacturing bolts using three machines, A, Band C. Of the total output, machine A is responsible for 25%, machine B for35% and machine C for the rest. It is known from previous experience with themachines that 5% of the output from machine A is defective, 4% from machineB and 2% from machine C. A bolt is chosen at random from the productionline and found to be defective. What is the probability that it came from

(a) machine A (b) machine B (c) machine C?

Let D={bolt is defective}, A={bolt is from machine A}, B={bolt is from machine B}, C={boltis from machine C}. We know that P(A) = 0.25, P(B) = 0.35 and P(C) = 0.4. AlsoP(D|A) = 0.05, P(D|B) = 0.04, P(D|C) = 0.02. A statement of Bayes’ Theorem for threeevents A,B and C is

P(A|D) = P(D\A)P(A) P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) = 0.05 x 0.025

0.5 × 0.25 + 0.04 × 0.35 + 0.02 × 0.4 =0.362P(B|D) = 0.04 x 0.35 = 0.406

0.5 × 0.25 + 0.04 × 0.35 + 0.02 × 0.4P(C|D) = 0.02 x0.4 = 0.232

0.5 × 0.25 + 0.04 × 0.35 + 0.02 × 0.40.6

4. Obtain the sample space of an experiment that consists of a ‘fair’ coin being tossed fourtimes. Consider the following events:A is the event ‘all four results are the same.’B is the event ‘exactly one Head occurs.’C is the event ‘at least two Heads occur.’Show that P(A) + P(B) + P(C) = 1716 and explain why P(A) + P(B) + P(C) > 1. P(A) =2/16 , P(B) = 4/16, P(C) = 11/16 , P(A) + P(B) + P(C) = 17/16

5. Machines A and B produce 10% and 90% respectively of the production of a componentintended for the motor industry. From experience, it is known that the probability thatmachine A produces a defective component is 0.01 while the probability that machine Bproduces a defective component is 0.05. If a component is selected at random from a day’sproduction and is found to be defective, find the probability that it was made by(a) machine A;(b) machine B. Let A = {item from machine A}, B = {item from machine B}, D = {item is defective}.

P(A|D) =P(D|A)P(A)/P(D|A)P(A) + P(D|B)P(B) =0.01 × 0.1/0.01 × 0.1 + 0.05 × 0.9 = 0.02

Similarly P(B|D) = 0.98

6. A garage mechanic keeps a box of good springs to use as replacements on customers cars.The box contains 5 springs. A colleague, thinking that the springs are for scrap, tossesthree faulty springs into the box. The mechanic picks two springs out of the box whileservicing a car. Find the probability that:(a) the first spring drawn is faulty;(b) the second spring drawn is faulty.

Let A ={first spring chosen is faulty}, B ={second spring chosen is faulty}(a) P(A) = 3/8 (b) P(B) = P(B|A)P(A) + P(B|A )P(A ) =2/7 x3/8 +3/7x 5/8 = 21/56 = 3/8

7. Suppose that we have two bags each containing black and white balls. One bag contains three times as many white balls as blacks. The other bag contains three times as many black balls as white. Suppose we choose one of these bags at random. For this bag we select five balls at random, replacing each ball after it has been selected. The result is that we find 4 white balls and one black. What is the probability that we were using the bag with mainly white balls?

P(Bla1)= (5/1) (3/4)4(1/4)1= 405/1024 similarly P(Bla2) = (5/1)(1/4)4(3/4)1= 15/1024

hence P(a1lB)=(405/1024 )/405/1024 + 15/1024 = 405 /420=0.964

8. A manufacturer claims that its drug test will detect steroid use (that is, show positive for an athlete who uses steroids) 95% of the time. What the company does not tell you is that 15% of all steroid-free individuals also test positive (the false positive rate). 10% of the rugby team members use steroids. Take E = the event that a rugby team member tests positive F = the event that a rugby team member uses steroids

n the example we just considered, P(E|F) = 0.95 P(E|F') = 0.15 P(F) = 0.1 P(F') = 0.9

P(E|F)= (0.95)(0.1)/(0.95)(0.1)+ (0.15)(0.9) =0.4130

9. The consumers Electronics Company is considering marketing a new model of Television set.In the past, 40% of the Television sets introduced by the company have been successful and 60% have been unsuccessful. Before introducing the Television set to the market place , the marketing research department conducts an extensive study and releases a report, either favorable or unfavorable. In the past 80% of the successful Television sets had received a favorable report.For the new model of Television set under consideration, the marketing research department has issued a favorable report. What is the probability that the Television set will be successful?

Let; event S = successful Television set event F = favorable report event S’ = unsuccessful Television set event F’ = unfavorable report P(S)=0.40 P(F\S)=0.80 P(S’) =0.60 P(F\S’) =0.30

 P(F\S) = P(F\S) P(S)/ P(F\S) P(S) + P(F\S’) P(S’)

= 0.80 x 0.40 / 0.80 x 0.40 + 0.60 x 0.30 = 0.64

10. The probability that a person has a certain disease is 0.03, medical diagnostic test are available to determine whether the person actually has the disease. If the disease is actually present, probability that the medical diagnostic test will give a positive result (indicating that the disease is present ) is 0.90. If the disease is not actually present, the probability of a positive test result (indicating that the disease is present ) is 0.02. Suppose that the medical diagnostic test has given a positive result (indicating that the disease is present ). What is the probability that the disease is present is actually present? What is the probability of a positive test result?

Let event D = has disease event T = test is positive event D’ = does not have disease event T’ = test is negative P(D) = 0.03 P(T\D) = 0.90 P(D’) =0.97 P(T\D’) =0.02

P(T\D) = P(T\D) x P(D) / P(T\D) x P(D) + P(T\D’) x P(D’) = ( 0.90 )( 0.03) / ( 0.90 )( 0.03) + ( 0.02 ) x (0.97) = 0.582